'*** PROGRAM: Compare 2 numbers using both min/max functions;
' then, output which number is max/and, which is min.
'*** declare variables...
min = 0
max = 0
number1 = 342
number2 = 256
'*** main program...
CLS '...(CL)ear the Output (S)creen
PRINT "Minimum = "; findMin(number1, number2)
PRINT "Maximum = "; findMax(number1, number2)
END '...END of program/halt program code execution
FUNCTION findMax (num1, num2)
answer = 0
IF num1 > num2 THEN answer = num1 ELSE answer = num2
findMax = answer
END FUNCTION
FUNCTION findMin (num1, num2)
answer = 0
IF num1 < num2 THEN answer = num1 ELSE answer = num2
findMin = answer
END FUNCTION
---program output...
Minimum: 256
Maximum: 342
Find the maximum and minimum values that the function can take over all the values in the domain for the input. The range is the maximum minus the minimum.
y=2x2-3x2-12x+5=0
Set the first derivative of the function equal to zero, and solve for the variable.
The general procedure is to find the function's derivative, and then solve for (derivative of the function) = 0. Each of these solutions may be a local maximum or minimum - or none. Further analysis is required. A local maximum or minimum may also occur at points where the derivative is undefined, as well as at the function's endpoints (assuming it is only defined for a certain range, for example, from 0 to 10).
the maximum or minimum value of a continuous function on a set.
By taking the derivative of the function. At the maximum or minimum of a function, the derivative is zero, or doesn't exist. And end-point of the domain where the function is defined may also be a maximum or minimum.
Find the maximum and minimum values that the function can take over all the values in the domain for the input. The range is the maximum minus the minimum.
Addition is the maximum or minimum function in math.
You cannot. The function f(x) = x2 + 1 has no real zeros. But it does have a minimum.
In Calculus, to find the maximum and minimum value, you first take the derivative of the function then find the zeroes or the roots of it. Once you have the roots, you can just simply plug in the x value to the original function where y is the maximum or minimum value. To know if its a maximum or minimum value, simply do your number line to check. the x and y are now your max/min points/ coordinates.
The minimum is the vertex which in this case is 0,0 or the origin. There isn't a maximum.....
y=2x2-3x2-12x+5=0
Set the first derivative of the function equal to zero, and solve for the variable.
To determine the maximum displacement, you need to calculate the peak value of the displacement function. This is done by finding the extreme values (maximum or minimum) of the function, typically by taking the derivative and setting it to zero to find critical points. Once you have these critical points, evaluate the function at those points to find the maximum displacement.
The general procedure is to find the function's derivative, and then solve for (derivative of the function) = 0. Each of these solutions may be a local maximum or minimum - or none. Further analysis is required. A local maximum or minimum may also occur at points where the derivative is undefined, as well as at the function's endpoints (assuming it is only defined for a certain range, for example, from 0 to 10).
Assuming the standard x and y axes, the range is the maximum value of y minus minimum value of y; and the domain is the maximum value of x minus minimum value of x.
In theory you can go down the differentiation route but because it is a quadratic, there is a simpler solution. The general form of a quadratic equation is y = ax2 + bx + c If a > 0 then the quadratic has a minimum If a < 0 then the quadratic has a maximum [and if a = 0 it is not a quadratic!] The maximum or minimum is attained when x = -b/2a and you evaluate y = ax2 + bx + c at this value of x to find the maximum or minimum value of the quadratic.